# HW3Sol - with n-1 degrees of freedom So for the desired...

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Homework 3 – solution 1. See the spreadsheet. The average observed biases we found to be : for n = 5, 1.83 and - 1 . 88, for a and b , respectively. For n = 10, they were 0.66 and - 1 . 13. For n = 50, they were found to be 0.30 and - 0 . 18, respectively. We see that as n increases the observed average biases decrease, as should be the case for any MLE. Note: your numbers will obviously be diﬀerent. 2. Using CLT, we obtain ¯ X N (5 . 00 , 0 . 05 20 ). So the probability can be found as Pr(4 . 98 < ¯ X < 5 . 02) = Φ ˆ 5 . 02 - 5 . 00 0 . 05 / 20 ! - Φ ˆ 4 . 98 - 5 . 00 0 . 05 / 20 ! = Φ(1 . 79) - Φ( - 1 . 79) = 0 . 93 . 3. We know that the statistic X 2 = ( n - 1) S 2 σ 2 has a chi-square distribution
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Unformatted text preview: with n-1 degrees of freedom. So, for the desired probability, we can write. Pr(0 . 03 < S < . 07) = Pr(0 . 0009 < S 2 < . 0049) = Pr â€¡ . 0009 Â· 19 . 0025 < X 2 < . 0049 Â· 19 . 0025 Â· = Pr(6 . 84 < X 2 < 37 . 24) = 0 . 995-. 007 = 0 . 988 . 4. We know that Ë† P âˆ¼ N Â± p, q p (1-p ) n Â¶ . So for the probability in question, we obtain Pr( Ë† P < . 60) = Î¦ . 60-. 70 q . 70 Â· . 30 200 = Î¦(-3 . 09) = 0 . 001 . 1...
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